A tightly wound long solenoid carries a curre | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The time period $T$ of a charged particle moving in a uniform magnetic field $B$ is given by $T = \frac{2\pi m}{qB}$.For a long solenoid,the magne

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(A) The time period $T$ of a charged particle moving in a uniform magnetic field $B$ is given by $T = \frac{2\pi m}{qB}$.For a long solenoid,the magnetic field inside is $B = \mu_0 n I$,where $n$ is the number of turns per metre and $I$ is the current.Substituting $B$ into the time period formula: $T = \frac{2\pi m}{q \mu_0 n I}$.Rearranging for $n$: $n = \frac{2\pi m}{q \mu_0 I T}$.Given values: $m = 9 	imes 10^{-31} 	ext{ kg}$,$q = 1.6 	imes 10^{-19} 	ext{ C}$,$I = 1.5 	ext{ A}$,$T = 75 	imes 10^{-9} 	ext{ s}$,and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ N/A}^2$.Substituting these values:$n = \frac{2\pi 	imes 9 	imes 10^{-31}}{1.6 	imes 10^{-19} 	imes 4\pi 	imes 10^{-7} 	imes 1.5 	imes 75 	imes 10^{-9}}$.$n = \frac{18\pi 	imes 10^{-31}}{9.6\pi 	imes 10^{-19} 	imes 10^{-7} 	imes 10^{-9} 	imes 75}$.$n = \frac{18 	imes 10^{-31}}{9.6 	imes 75 	imes 10^{-35}} = \frac{18 	imes 10^4}{720} = \frac{180000}{720} = 250 	ext{ turns/m}$.

Similar Questions

The magnetic induction along the axis of a toroidal solenoid is independent of

$A$ toroid has a number of turns per unit length $n$ and current $i$. What is the magnetic field inside the toroid?

Two coaxial solenoids of different radii carry current $I$ in the same direction. Let $\overrightarrow{F_1}$ be the magnetic force on the inner solenoid due to the outer one and $\overrightarrow{F_2}$ be the magnetic force on the outer solenoid due to the inner one. Then

Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid decreases.
Reason : Magnetic field energy density is proportional to the square of the current.

The number of turns per unit length of a long solenoid is $10 \, \text{turns/cm}$. If its average radius is $5 \, \text{cm}$ and it carries a current of $10 \, \text{A}$, then the ratio of flux densities obtained at the centre and at the end on the axis will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module